Feynman's QED and photon reflection

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A few days ago I started reading through Feynman's QED. It is a very good book, I am very much enjoying it.

There's something I don't understand. When Feynman starts to talk about why light is reflected such that the angle of incidence equals the angle of reflection, he gives an example where the light source is the same distance from the mirror as the photomultiplier (for simplicity). This gives the final result of the probability amplitude. This is mostly due to the middle part of the mirror, when the angles of the probability "vectors" differ so slightly that they add up greatly, where as later they matter less and less to the final vector.

This is important for every explanation (that I have thus far) in the book. The fastest path for the photon is the most likely.

By taking certain parts away from the mirror, he demonstrated that what was going on really was reflection. I understood this.

However, when I was thinking about this later, I tried to think of a scenario where the photomultiplier is further or closer to the mirror than the light source. It came to my attention that the fastest path that a photon could take would not be such that the angle of incidence were to equal the angle of reflection, but such that the length of the line from the light source to the mirror would equal the length of the line from the photomultiplier to the mirror. When the photomultiplier and the light source are at the same distance from the mirror, then those lines are such that the angle of incidence were to equal the angle of reflection, as in the book.

However, light should always travel such that the angle of incidence were to equal the angle of reflection. There's something I'm not grasping here.

Thanks
 

vanhees71

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You should not think about photons in terms of classical particles. If you put the photo detector at a place, where a classical ray would not enter it, what you measure is (a pretty) small probability that a photon is scattered into the detector although in ray optics there would be no light.

Feynman uses the path-integral formulation of QED in this book, and what he shows is that in the classical limit, formally given by the leading-order eikonal approximation of the path integral (stationary-phase approximation) the photon travels most probably along the path of a light ray. This is of course all correct, but one should not overstress the analogy between photons and classical particles. It's not even possible to define a position observable for a photon in the usual sense, because photons are massless.
 
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You should not think about photons in terms of classical particles. If you put the photo detector at a place, where a classical ray would not enter it, what you measure is (a pretty) small probability that a photon is scattered into the detector although in ray optics there would be no light.

Feynman uses the path-integral formulation of QED in this book, and what he shows is that in the classical limit, formally given by the leading-order eikonal approximation of the path integral (stationary-phase approximation) the photon travels most probably along the path of a light ray. This is of course all correct, but one should not overstress the analogy between photons and classical particles. It's not even possible to define a position observable for a photon in the usual sense, because photons are massless.
Right. Feynman touches on this with a few example. But that doesn't help me to understand why the angle of incidence = angle of reflection
 

vanhees71

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The classical limit within the path-integral formalism is calculated using the method of stationary phase. Rougly speeking, in the path integral you integrate the phase factor [tex]\exp(\ii S[A]/\hbar)[/tex] over all field configurations, allowed by the boundary conditions given by the setup of the environment, in your case an ideally conducting plane metal to describe the mirror. Now, the stationary phase is given by the field configuration, for which the classical action [tex]S[A][/tex] becomes stationary. This, however, leads precisely to the classical Maxwell equations for the em. field, and solving it under the constraints of the boundary conditions leads to the usual rule of reflection for the wave vectors (outgoing angle=incoming angle wrt. to the normal of the mirror surface).

Quantum mechanically this means that field configurations around this classical field configuration leads to constructive interference of the integral while other field configurations almost completely annihilate each other because the phase changes quite rapidly.

Using further the eikonal approximation for the classical em. wave, you see that a light ray, is reflected in the usual way, at least in regions, far from the mirror compared to the wave length of the incoming (and outgoing!) wave.
 

DrDu

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I think this is a simple question of ray optics and that the statement that the shortest (or fastest path is the one with equal length of the path from the source and detector, respectively, to the mirror is wrong. The standard argument as to why the path with incidence angle= reflection angle is shortest is that upon reflection of the reflected path, one gets a straight line, which is the shortest connection of two points.
How did you get to your conclusion?
 

vanhees71

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Fermat's principle tells you that the time the light travels along the total path of the incoming and reflected ray is minimal at fixed points of the source and of observation. In ray optics there's no light except on the rays, i.e., here the qestion is, where the incoming ray has to hit the mirror such that the total path is minimal.

You can of course assume that the light rays are straight lines. Then the extremal problem becomes a simple problem to find the minimal path by varying the point of incidence of the incoming ray with the mirror. What you get out is the Law of Reflection: The angle of the incoming ray with the normal to the mirror is the same as the angle of the outgoing ray.

Of course, you have to solve the extremal problem under the constraint that the first ray hits the mirror. Without that restriction you get of course a straight line connecting the two given fixed points.
 

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